{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 1. 无穷小的比较"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "假设 $\\alpha$ 与 $\\beta$ 都是同一个自变量的变化过程中的无穷小，且 $\\alpha$ $\\not=$ 0\n",
    "- 如果 **$\\lim\\limits_{}\\frac{\\beta}{\\alpha} = 0$**, 那么说 $\\beta$ 是比 $\\alpha$ **高阶** 的无穷小，记作：$\\beta$ = o($\\alpha$)\n",
    "- 如果 **$\\lim\\limits_{}\\frac{\\beta}{\\alpha} = \\infty$**, 那么说 $\\beta$ 是比 $\\alpha$ **低阶** 的无穷小\n",
    "- 如果 **$\\lim\\limits_{}\\frac{\\beta}{\\alpha} = c \\not 0$**, 那么说 $\\beta$ 是比 $\\alpha$ **同阶** 的无穷小\n",
    "- 如果 **$\\lim\\limits_{}\\frac{\\beta}{\\alpha} = 1$**, 那么说 $\\beta$ 是比 $\\alpha$ **等价** 的无穷小"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 2.求 $\\lim\\limits_{x\\rightarrow0}\\frac{sinx}{3x+x^3}$,并用 python 验证编程"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "解： 因为当 x = 0,$\\lim\\limits_{x\\rightarrow0}\\frac{sinx}{x}$ = 1,所以 sinx 与 x 等价无穷小\n",
    "    因此 $$\\lim\\limits_{x\\rightarrow0}\\frac{sinx}{3x + x^3} = \\lim\\limits_{x\\rightarrow0}\\frac{x}{3x + x^3} = \\lim\\limits_{x\\rightarrow0}\\frac{1}{3 + x^2} = \\frac{1}{3}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### python 验证"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1/3\n"
     ]
    }
   ],
   "source": [
    "import sympy\n",
    "import numpy as np\n",
    "\n",
    "x = sympy.Symbol(\"x\")\n",
    "f = sympy.sin(x)/(3*x + x**3)\n",
    "s = sympy.limit(f, x, 0)\n",
    "print(s)"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.8.5"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 4
}
